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Torques of equal magnitude are applied to a hollow cylinder and a solid sphere , both having the same mass and radius . The cylinder is free to rotate about its standard axis of symmetry , and the sphere is free to rotate about an axis passing through its centre . Which of the two will acquire a greater angular speed after a given time . |
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Answer» Solution :Let M and R be mass and radius of the hollow cylinder and the solid SPHERE , then Moment of INERTIA of the hollow cylinder about its axis of SYMMETRY , `I_(1) = MR^(2)` Moment of inertia of the solid sphere about its axis through its centre `I_(2) = (2)/(5) MR^(2)` Torque `TAU = I alpha` where `alpha` is angular acceleration . `I_(1) alpha_(1) = I_(2) alpha_(2)` `(alpha_(2))/(alpha_(1)) = (I_(1))/(I_(2)) = (MR^(2))/((2)/(5) MR^(2)) = (5)/(2)` `alpha_(2) = 2.5 alpha_(1) or alpha_(2) gt alpha_(1)` From `omega = omega_(0) + alpha t` , we find that for given `omega_(0)` and `t , omega_(2) gt omega_(1)` , angular speed of solid sphere will be greater than the angular speed of hollow cylinder . |
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